[AlphaScent]

So this is a question from my undergraduate organic lab that I am proctoring (teachers assistant) for the semester.  This question comes from there book and want to be sure I understand what is going on.  Some stuff it has been a while.

So,

A mixture of toluene (bp 110.8 C) and water is steam distilled.  Visual inspection of the distillate reveals that there is a greater volume of toluene than water present, yet water (bp: 100 C) has the higher vapor pressure. 
Explain the observation.

My thoughts:

It is too easy to say that it forms an azeotrope.  The chapter doesnt discuss azetropes.  Though the chapter before does.  I wasnt there last semester I do not know.  My thoughts are to just say that it does not conform to Raoults law and therefore the mole fraction does not matter.  It forms a constant boiling mixture.  It behaves as one compound.

What else is there to say/ Is there a better way to explain it?

This may be in the wrong spot.  MOd feel free to move it if it is.

[TheUnassuming]

Lit has the water/toluene azeotrope at ~80% by mass... so the azeotrope explanation is perhaps right in a way, just not detailed enough of an explanation? 
If they are covering Raoults law in this chapter the answer probably revolves around that.  I would just say that the toluene/water mix is not an ideal gas/liquid so the forces between the two liquids aren't uniform, making it possible for the mol fraction in the gas phase to be different than the liquid.  In the toluene/water case this gives the solution a positive deviation from Raoults law.  I don't know that there is really much else to say, especially for an undergrad organic lab. 

[AlphaScent]

I am over thinking this I guess.  I wonder if any of them will even come up with azeotrope.

Thanks Unassuming

Cheers

[TheUnassuming]

I wonder if any of them will even come up with azeotrope.

haha... depends on the composition of the class.  Sophomore organic class I would expect a few of them to get it since there should be a few chem majors in there.